编写一个在1，2，…，9（顺序不能变）数字之间插入+或-或什么都不插入，使得计算结果总是100的程序。并输出所有的可能性。

/*
 * 可以看成是在数字中间插入运算符。插入 + - 或空位。那么每个空位有3种选择，加号，减号或者空。
 * 一共8个位置，共有3的8次方个情况。每个情况都试一次，找出结果为100的组合。
 * 
 */
public class ResultOneHundred {
	
	public static void main(String args[]){
		String num[] = new String[17];
		int count1 = 1;
		for (int i = 0; i < num.length; i++) {
			if(i%2 == 0){
				num[i] = count1 + "";
				count1++;
			}
			else{
				num[i] = "";
			}
		}//准备好特定格式的数组
		
		System.out.println("Start");
		get100(num, 1, 16, num);
		System.out.println("End");
	}
	
	/**
	 * 找出结果为100的表达式
	 * @param data1
	 * @param count
	 * @param times
	 * @param data2
	 */
	static void get100(String data1[], int count,int times,String data2[]){
		if(count >= times){
			if (calculate100(data2) == 100){
				
				for (int j = 0; j < data2.length; j++) {
					System.out.print(data2[j]);
				}
				System.out.println( " = 100");
			}
			return;

		}
		
		String temp1[] = new String[data1.length];
		String temp2[] = new String[data1.length];
		String temp3[] = new String[data1.length];
		String add = "+";
		String minus = "-";
		String empty = "";
		copyStringArray(temp1, data2);
		copyStringArray(temp2, data2);
		copyStringArray(temp3, data2);
		
		temp1[count] = add;
		temp2[count] = minus;
		temp3[count] = empty;
		
		temp1[count+1] = data1[count+1];
		temp2[count+1] = data1[count+1];
		temp3[count+1] = data1[count+1];
		count+=2;
		
		get100(data1, count, times, temp1);
		get100(data1, count, times, temp2);
		get100(data1, count, times, temp3);
	}
	/**
	 * 复制字符数组的方法，从from[]复制到to[]
	 * @param to
	 * @param from
	 */
	static void copyStringArray(String to[], String from[]){
		for (int i = 0; i < from.length ; i++) {
			to[i] = from[i];
		}
	}
	/**
	 * 求和方法，输入一个String[]运算式子
	 * 数组要写成指定的格式,大于10的数之间要用""隔开。
	 * @see ["1","","4"<span style="font-family: Arial, Helvetica, sans-serif;">,""</span>,"7","+","7"]
	 * @param cal
	 * @return 表达式的和
	 */
	static int calculate100(String cal[]){
		int sum = 0;
		int bitCount = 0;
		int tempNum = 0;
		for (int i = cal.length - 1; i >= 0; i--) {
			
			if (cal[i] == "+") {
				sum+=tempNum;
				tempNum = 0;
				bitCount = 0;
			}
			else if (cal[i] == "-"){
				sum-=tempNum;
				tempNum = 0;
				bitCount = 0;
			}
			else if (cal[i] == ""){
				bitCount++;
			}
			else{
				tempNum+=Math.pow(10, bitCount) * Integer.parseInt(cal[i]);
			}
		}
		return sum+=tempNum;
	}
}

Start
1+2+3-4+5+6+78+9 = 100
1+2+34-5+67-8+9 = 100
1+23-4+5+6+78-9 = 100
1+23-4+56+7+8+9 = 100
12+3+4+5-6-7+89 = 100
12+3-4+5+67+8+9 = 100
12-3-4+5-6+7+89 = 100
123+4-5+67-89 = 100
123+45-67+8-9 = 100
123-4-5-6-7+8-9 = 100
123-45-67+89 = 100
End